Monotone Convergence

Monotone Convergence Theorem (MCT). A sequence is monotonically increasing (or just increasing) if for all n. Now the theorem says: an increasing sequence with an upper bound is convergent.

Proof.

Let L = sup{a1, a2, … }, which exists by completeness of R. Thus, each an ≤ L. On the other hand, for each ε>0, since L-ε is not an upper bound of the {an}, we can find aN > L-ε. Hence for all n > N, we have an ≥ aN > L-ε while an ≤ L which gives:

♦

In practice, the monotonic convergence theorem often tells us a limit exists without giving any hint on what it might be. In some cases, it may be derivable.

Answer. We shall prove that by induction. For n=0 this is obvious. Suppose we have . For the next iteration, we have:

This shows that which completes the induction. Next, note that since xn > 0. Hence, (xn) is an increasing sequence with an upper bound and must converge to some L.

From , the LHS → L while the RHS → 6-(5/L). Equating and solving give L=1 or L=5. Since each xn ≥ 2, the limit can’t be 1. So L=5. ♦

Example 2. Prove that converges.

Answer. Let an be the n-th partial sum. Since n! ≥ 2n-1, we have . But converges by sum of geometric series. Hence, an is an increasing sequence with an upper bound. By MCT, it converges. ♦

However, there’s no easy way to evaluate this sum.

Note: in proving MCT, we explicitly used the fact that R is complete. This property is critical since the theorem does not hold for Q. For example, we can easily have the sequence 1, 1.4, 1.41, 1.414, … which contains more and more significant figures of √2. We get an increasing sequence of rational numbers, with no limit in Q. On the other hand, since R is complete, there’s a limit in R.

This gives a good sense of what completeness means. We get a mental picture of “gaps” occurring in Q, so that a sequence of elements which get successively closer still may not converge. The next section will elucidate this even further.

Cauchy Sequences

We’ll define what it means for a sequence of numbers to get successively closer.

Definition. A Cauchy sequence is a sequence (an) where

for any ε>0, there exists N such that when m, n > N, we have |am – an| < ε.

Some results are quite easy to prove.

A convergent sequence is Cauchy.

A Cauchy sequence is bounded.

Proof.

For the first statement, suppose the sequence → L. For any ε>0, there exists N such that when n > N, we have |an – L| < ε/2. Hence, when m, n > N, we have . So the sequence is Cauchy.

For the second, pick ε=1. We get an N such that when m, n > N, we get |am – an| < 1. Fix some n>N and let B = |an|. It follows that for all m>N, |am| ≤ |am – an| + |an| < B+1. So the set of am for m > N is bounded. Since there’re only finitely many terms up to N, the whole sequence is bounded. ♦

Notice that the property of being a Cauchy sequence is inherent in the sequence and does not rely on the embedding. In short, a sequence of rational numbers is Cauchy regardless whether we consider it embedded in Q or in R. On the other hand, whether a sequence of rational numbers converges depends on whether the ambient space is Q or R.

Theorem. A Cauchy sequence of real numbers converges to a real number.

Proof. Suppose (an) is Cauchy. In particular, it’s bounded and we can let . Then since bn takes the sup of more elements than bn+1. Since (an) has a lower bound so does (bn) and by MCT, (bn) converges to some L.

It remains to show (an) → L. Let ε > 0. Find integer N such that:

when m, n > N, we have |am – an| < ε/2;

when n > N, we have |bn – L| < ε/2, or L-(ε/2) <bn < L+(ε/2).

Now since L-(ε/2) <bN+1, and sup is the minimum upper bound, L-(ε/2) is not an upper bound of so we can find n > N such that . Yet , which gives .

Hence, for all m > N, we have . ♦

Squeeze Theorem

The following theorem is simple but handy.

Squeeze Theorem. Let be three sequences with . Suppose an → L and cn → L. Then bn → L.

Proof. Let ε > 0. Pick N such that whenever n > N :

;

;

Hence when n > N, we have and , i.e. . ♦

Example 3. Prove that has a limit of 0.

Proof. Since sin(n) ≤ 1, we have 2sin(n) ≤ 2 and . Since 2/n2 → 0, the result follows from the squeeze theorem. ♦

The new sequence which is obtained via taking the mean of the first n terms is called the Cesàro mean. The above exercise tells us taking the Cesàro mean preserves the limit if it already exists. On the other hand, some non-converging sequences converge after taking the Cesàro mean. E.g. if an = (-1)n, then its Cesàro mean converges to 0 since the n-th term is bounded by [-1/n, +1/n], so it converges by the squeeze theorem.

Subsequences

A subsequence of a sequence (an) is a sequence of the form (bm), where

, for some increasing n(1) < n(2) < n(3) < …

Thus in the sequence an = (-1)n, the even terms form a subsequence (+1, +1, +1, …) with indices n(1)=2, n(2)=4, n(3)=6, … . The odd terms form a subsequence (-1, -1, -1, …) with indices n(k)=2k-1. Hence, a non-convergent sequence can have subsequences which converge to different values.

Theorem. Let (an) be a sequence.

If lim (an) = L, then every subsequence converges to L.

If every subsequence converges to the same value L, then lim (an) = L.

Proof.

For the first statement, suppose we have a subsequence (bm), where , n(1) < n(2) < n(3) < … . Let ε>0. There exists N such that when n>N, |an–L| < ε. Since n(m) ≥ m, whenever m>N, we also have . Hence the subsequence (bm) also converges to L.

For the second statement, suppose (an) doesn’t converge to L. Unwinding the definition, there exists ε>0 such that for each N, there exists n>N such that |an–L| ≥ ε. In short, there are infinitely many n for which |an–L| ≥ ε. If we pick this subsequence, it cannot possibly converge to L. ♦